/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A baseball pitcher throws a base... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball pitcher throws a baseball horizontally at a linear speed of \(42.5 \mathrm{~m} / \mathrm{s}\) (about 95 \(\mathrm{mi} / \mathrm{h}\) ). Before being caught, the baseball travels a horizontal distance of \(16.5 \mathrm{~m}\) and rotates through an angle of 49.0 rad. The baseball has a radius of \(3.67 \mathrm{~cm}\) and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

Short Answer

Expert verified
The tangential speed of a point on the equator of the baseball is approximately 4.63 m/s.

Step by step solution

01

Convert Radius to Meters

First, let's convert the radius of the baseball from centimeters to meters for consistent units. Given that the radius is \(3.67\, \text{cm}\), we convert it to meters: \(3.67\, \text{cm} \times \frac{1 \text{m}}{100 \text{cm}} = 0.0367\, \text{m}\).
02

Find the Angular Speed

Next, we need to calculate the angular speed \(\omega\) of the baseball. Angular speed is given by the formula \(\omega = \frac{\theta}{t}\), where \(\theta\) is the angular displacement and \(t\) is the time. We know \(\theta = 49.0\, \text{rad}\). The time of travel \(t\) can be found from the horizontal speed equation \(v = \frac{d}{t}\) where \(d\) is the horizontal distance \(d = 16.5\, \text{m}\) and \(v = 42.5\, \text{m/s}\). So, \(t = \frac{16.5}{42.5} = 0.388 \text{ s}\). Hence, the angular speed is \(\omega = \frac{49.0}{0.388} \approx 126.29\, \text{rad/s}\).
03

Calculate Tangential Speed

Finally, the tangential speed \(v_t\) of a point on the equator of the baseball can be determined using the formula \(v_t = \omega \times r\), where \(\omega\) is the angular speed and \(r\) is the radius of the baseball. Thus, substituting the values found, \(v_t = 126.29\, \text{rad/s} \times 0.0367\, \text{m} \approx 4.63\, \text{m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Speed
Angular speed is a measure of how quickly an object rotates or revolves relative to another point, often the center of a circular path. It is designated by the Greek letter \(\omega\). Angular speed is determined by dividing the total angle of rotation by the time it takes to cover that angle. The formula is:
  • \(\omega = \frac{\theta}{t}\)
where:
  • \(\omega\) is the angular speed in radians per second
  • \(\theta\) is the angular displacement in radians
  • \(t\) is the time in seconds
In our exercise, to find the angular speed, we calculated the time it takes for the baseball to travel the horizontal distance and used the given angular displacement. This allowed us to determine how fast the baseball was spinning around its axis, giving us an angular speed of approximately \(126.29\ \text{rad/s}\). This concept is crucial in understanding rotational motion.
Tangential Speed
Tangential speed is the speed of something moving along a circular path. It refers to how fast a point on a rotating object is moving in a direction tangent to the circle. The tangential speed can be found using the formula:
  • \(v_t = \omega \times r\)
where:
  • \(v_t\) is the tangential speed in meters per second
  • \(\omega\) is the angular speed
  • \(r\) is the radius of the circular path
In this baseball exercise, the tangential speed was calculated for a point on the "equator" of the baseball. This was achieved by multiplying the angular speed by the radius of the baseball. It resulted in a tangential speed of approximately \(4.63\ \text{m/s}\). This indicates how fast a point on the surface of the baseball is traveling as it spins.
Radius Conversion
Conversions are often necessary in physics to ensure that the units align for calculations. In the context of circular motion, the radius must be in meters if we're working in the metric system, as this helps maintain consistency in units such as meters per second for speed. To convert the radius from centimeters to meters, you should:
  • Use the conversion factor: \(1\ \text{m} = 100\ \text{cm}\)
For example, given a radius of \(3.67\ \text{cm}\):
  • \(3.67\ \text{cm} \times \frac{1 \text{m}}{100 \text{cm}} = 0.0367\ \text{m}\)
This was the first step in solving our exercise. Ensuring the radius is in meters allows for correct and straightforward application of formulas involving radius, such as when calculating tangential and angular speed.
Angular Displacement
Angular displacement is the measure of the angle through which an object rotates around a particular point or axis. It is usually measured in radians, which is the standard unit of angular displacement in the metric system.
  • Angular displacement represents the cumulative angle turned, unlike linear displacement which reflects distance covered.
In the exercise, the baseball rotates by an angle of \(49.0\ \text{rad}\) as it travels downfield. This angular displacement is crucial when calculating angular speed, as it describes the total rotation relative to a given point within a specific timeframe. Understanding angular displacement helps us comprehend how far an object has turned from its initial position.

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Most popular questions from this chapter

The drill bit of a variable-speed electric drill has a constant angular acceleration of \(2.50 \mathrm{rad} / \mathrm{s}^{2} .\) The initial angular speed of the bit is \(5.00 \mathrm{rad} / \mathrm{s}\). After \(4.00 \mathrm{~s},\) (a) what angle has the bit turned through and (b) what is the bit's angular speed?

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has a length of \(64 \mathrm{~cm}\) and is wound around the top at a spot where its radius is \(2.0 \mathrm{~cm}\). The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of \(+12 \mathrm{rad} / \mathrm{s}^{2}\). What is the final angular velocity of the top when the string is completely unwound?

A string trimmer is a tool for cutting grass and weeds; it utilizes a length of nylon "string" that rotates about an axis perpendicular to one end of the string. The string rotates at an angular speed of \(47 \mathrm{rev} / \mathrm{s},\) and its tip has a tangential speed of \(54 \mathrm{~m} / \mathrm{s}\). What is the length of the rotating string?

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of \(-4.00 \mathrm{rad} / \mathrm{s}^{2}\). Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of \(-25.0 \mathrm{rad} / \mathrm{s}\). While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

A 220 -kg speedboat is negotiating a circular turn (radius \(=32 \mathrm{~m}\) ) around a buoy. During the turn, the engine causes a net tangential force of magnitude \(550 \mathrm{~N}\) to be applied to the boat. The initial tangential speed of the boat going into the turn is \(5.0 \mathrm{~m} / \mathrm{s}\). (a) Find the tangential acceleration. (b) After the boat is \(2.0 \mathrm{~s}\) into the turn, find the centripetal acceleration.

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